Category: Chemical Engineering Math
"Published in Newark, California, USA"
A 50.0 g sample of calcium carbonate is reacted with 35.0 g of phosphoric acid. Calculate
(a) the number of grams of calcium phosphate that could be produced.
(b) the number of grams of excess reagent that will remain.
(Atomic Weights: Ca = 40, C = 12, O = 16, H = 1, and P = 31)
Solution:
The first thing that we need to do is to write the chemical equation of the reaction as follows
Balance the equation by putting the corresponding coefficients as follows
Molecular Weight of CaCO3:
Molecular Weight of H3PO4:
Molecular Weight of Ca3(PO4)2:
Molecular Weight of H2CO3:
Since the weight of the two reactants are given in the problem, then we need to find out which of them is limiting reagent. Limiting reagent is a substance that gives the least amount of product in the reaction. On the other hand, excess reagent is a substance that gives the excess amount of a substance in the reaction.
Weight of Ca3(PO4)2 from CaCO3:
Weight of Ca3(PO4)2 from H3PO4:
By comparing the weights of Ca3(PO4)2, CaCO3 will give us the least amount of product.
(a) Therefore, the limiting reagent is CaCO3 and the amount of Ca3(PO4)2 produced is
Since CaCO3 is the limiting reagent, then the amount of H3PO4 can be used in the reaction is
(b) Therefore, the excess amount of H3PO4 after the reaction is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
For the reaction
Calculate the potential or voltage if the [Cu+2] is 1 x 10-4 and the [Al+3] is 1 x 10-3.
Solution:
From the given chemical equation above
Aluminum lost six electrons and copper gained six electrons. The chemical equation is already balanced including the ions.
From Table of Common Standard Reduction Potentials for Al, E0 = 1.662 V since aluminum lost three electrons and oxidation takes place in the reaction. If you multiply any of half-reactions with a coefficient, then the value of reduction potential is still the same. We don't multiply the reduction potential with any coefficients and we change only the sign if the half-reaction is reversed.
Again, from Table of Common Standard Reduction Potentials for Cu, E0 = 0.3419 V since copper gained two electrons and reduction takes place in the reaction.
The reduction potential of the reaction is the sum of the reduction potential of two half-reactions. Hence, the reduction potential is
If the concentration of metal ions are given, then we can calculate the voltage of the reaction by using Nernst Equation as follows
where n is the number of electrons lost or gained in the reaction. Hence, the voltage for the reaction is
For any whole or pure metals in half-reactions, their concentration is 1 M. In this case, the concentration of Al and Cu is 1 M. Therefore, the voltage for the reaction is
Category: Chemical Engineering Math
"Published in Newark, California, USA"
Calculate the voltage of the cell Fe; Fe+2 ║H+; H2 if the iron half cell is at standard conditions but the H+ ion concentration is 0.001 M.
Solution:
The half-cell reaction for Fe is
and it is an oxidation reaction because iron lost two electrons. From Table of Common Standard Reduction Potentials, E0 = 0.447 V. If you reverse the half-cell reaction, the sign of reduction potential will change. Since Fe+2 is at standard condition, then the concentration is 1 M. The concentration of any whole or pure metal in the half-cell reaction is 1 M.
If the concentration of any metal ion is given, then we can calculate the voltage of half-cell reaction by using Nernst Equation as follows
where n is the number of electrons lost or gained in half-cell reaction. Hence, the voltage for Fe is
The half-cell reaction for H is
and it is a reduction reaction because hydrogen gained two electrons. From Table of Common Standard Reduction Potentials, E0 = 0 V. If you reverse the half-cell reaction, the reduction potential is still the same because it is 0 V. If the concentration of H+ is 0.001M, then the voltage for H by Nernst Equation is
Therefore, the voltage of the cell is
